Find the value of nnn for which the function f(x)=xnf(x) = x^nf(x)=xn satisfies the identity x2f′′(x)=2f(x)x^2 f''(x) = 2f(x)x2f′′(x)=2f(x).
n=1n=1n=1
n=2n=2n=2
n=3n=3n=3
n=4n=4n=4